6
$\begingroup$

To be very-very simple:

The very main setting of my world is a very special and very small galaxy, the galactic core of which is a quasar - with the "iconic" light beam in the middle, serving a special and iconic object for the cultures living within the galaxy.

According to a Redditor, these are extremely bright objects, with the possibility of providing enough brightness to eliminate night, even from tens of thousands of light years.

However, I decided to choose a very small "population": less than 1 million stars. If a quasar is so strong, then it would be an extremely sparse galaxy and I need to do something about it.

What is the estimated distance from a small quasar (if it's a thing at all) where brightness would not interfere normal day-night cycles?

$\endgroup$
  • $\begingroup$ you could put it on the same plane as the planets orbit and use a slightly weaker sun then you would have a day night cycle but the night part of the cycle will vary a lot more than is normal. you could get seasons without an axial tilt then as well. $\endgroup$ – John Jan 7 '17 at 5:10
  • 1
    $\begingroup$ You know that quasars are also old, associated with early galaxy formation. Today, the same place would be a normal galaxy. So…what planets? Other than brown dwarfs, no planets were forming because heavier elements had not formed yet. $\endgroup$ – JDługosz Jan 7 '17 at 6:41
  • $\begingroup$ (a) Is the quasar continuous? I don't think that's a thing. (b) Do you still want the quasar to be visible in the night sky? $\endgroup$ – SIGSTACKFAULT Jan 9 '17 at 19:20
  • $\begingroup$ The very small population does not go well with current observational evidence. A million stars doesn't even make a good globular cluster, much less a galaxy. Even dwarf galaxies like the small magelan cloud are estimated to have billions of solar masses. $\endgroup$ – Durandal Jan 9 '17 at 20:21
  • 1
    $\begingroup$ @Katamori I think I explained that poorly. Put a different way, the majority of a quasar's energy is emitted away from the galactic plane, and most planets orbiting in the galactic plane shouldn't receive much radiation. In other words, unless the planets have wacky orbits, they should be relatively fine, and the brunt of the jets will miss them entirely. $\endgroup$ – HDE 226868 Aug 9 '18 at 12:48
3
$\begingroup$

The size and luminosity of quasars varies by a rather large degree, since you did not bother to include specifications, I'll supply my own based on the average:

Quasars emit a light as bright as that of one trillion stars and they have a radius of about 90 AU (0.00142313 light-years).

Light travelling through the vacuum is subject to the inverse-square law which means that the perceived intensity of the light is inversely proportional to the square of the distance from the light source: $I = 1/r^2$

I have no idea how dark you want nights on your planet, all you say is you want there to be a clear day-night cycle, so I'm going to say you don't want the perceived brightness of your quasar to be anymore then a tenth of that of a average star. That way, you will still indeed have day-night cycles but the quasar will still be a very impressive sight in the night sky (certainly sufficient to be an iconic object). In other words you want the perceived intensity to be only 1/10 trillionth of what it is at the source (which is the surface of the quasar at a radius of 90 AU from its center).

So let's plug in the numbers.

1/r^2 = 1/10,000,000,000,000

Now if we solve for $r$ we get about 3,162,277. The unit as I've previously mentioned is 90 AU. When we convert to whole light-years we get 4500 light-years.

Now considering that 1/10th the brightness of an average star is still very large and that the Milky Way galaxy which happens to contain anywhere from 100-400 billion stars, only has an estimated radius of 100,000 light-years than we can see that 4500 light-years is rather big for a galaxy only containing a million stars. If you were to increase the number of stars then this would become feasible.


Note: Thank you to Adrian Colomitchi for pointing out the mistake I had previously made.

$\endgroup$
  • $\begingroup$ At the end, I highlighted that I'm satisfied with an answer that tells the estimated distance from a small quasar (if it's a thing at all) - but an average value is a sufficient answer too. This one definitely shows that I have to apply some "handwave" and/or non-scientific "trickery" to decrease these disadvantages and that's what I wanted to know. $\endgroup$ – Katamori Jan 7 '17 at 2:46
  • 1
    $\begingroup$ Oh sorry about that I didn't notice, I'd really have prefered not to have given only a "sufficient" answer. However the point still stands as average quasars are already very small compared to the few absolutely monstrous quasars that exist(there don't seem to be any uniquely tiny quasars because of the very process by which they are created). Further more even if we were to change the size of the quasar to the very smallest possible, we'd still arrive at a conclusion within the same order of magnitude. $\endgroup$ – AngelPray Jan 7 '17 at 3:00
  • $\begingroup$ Average quasars may be bright, smaller ones (or in periods of dimmed activity) seem to be less aggressive - see 3C 273. $\endgroup$ – Adrian Colomitchi Jan 7 '17 at 4:25
  • 1
    $\begingroup$ 'if anyone could enlighten me' Huh, got it! After a (relative short) while, here comes the light Quote: The luminosity of some quasars changes rapidly in the optical range and even more rapidly in the X-ray range. Because these changes occur very rapidly they define an upper limit on the volume of a quasar; quasars are not much larger than the Solar System.[8]. Seems that you made a bad assumption with "and they have a radius of about 0.75 lightyears." $\endgroup$ – Adrian Colomitchi Jan 7 '17 at 5:56
  • 1
    $\begingroup$ @Katamori if it is still too bright, just put some nebula or similar object between your planet and the quasar. We also cannot directly see the centre of Milky Way, only some bright "windows". Then the brightness can be scaled down and you will be able to put there even some even better looking scenery around that shiny thingie :) Also that means you can be more near to the object. $\endgroup$ – Antoine Hejlík Jan 9 '17 at 9:30
4
$\begingroup$

Something that looks in disagreement with @AngelPray's answer - the 3C 273

if it were only as distant as Pollux (~10 parsecs) it would appear nearly as bright in the sky as the Sun.

10psc is roughly 33 ly. Inverse square law says if it would be 10x farther - i.e. 330 ly - the brightness will be 1/100 of Sun's brightness.

At 3300 ly (that's 1/10 the diameter of the Milky Way's size, 1/5 of its radius), the brightness would be 0.1% that of the Sun. This is in the same ballpark as the difference in brightness between the Sun and the Moon.

$\endgroup$
  • $\begingroup$ Wait... There must be a problem here. How in the world could it only be almost as bright as the sun at just 33 lightyears away? 3C 273 is 4 trillion times brighter then the sun... Honestly, I don't know what's incorrect, just that something must be. $\endgroup$ – AngelPray Jan 7 '17 at 5:05
  • $\begingroup$ Might have to do with the logarithmic perception of the brightness? $\endgroup$ – Adrian Colomitchi Jan 7 '17 at 5:10
  • $\begingroup$ After doing some digging, apparently you are absolutely correct. That is, with a galaxy the size of our own and not one with only a million stars. Sadly I still have no idea where in my calculations I went wrong. $\endgroup$ – AngelPray Jan 7 '17 at 5:25
  • $\begingroup$ @AngelPray - why would the size of the galaxy matter? I can imagine a super-hyper-gigantic massive blackhole just devouring the last remnants of its accretion disk formed by the inner/mid-range dist stars, over some time of 100ky, letting only the distant million stars surviving - at least for some million years yet. Intelligent life, capable to appreciate the beauty of the stellar carnage need less than 100ky to go from stone to bronze to iron and beyond post-information era (or even the post-truth one) $\endgroup$ – Adrian Colomitchi Jan 7 '17 at 5:40
  • $\begingroup$ That's not how blackholes behave. They don't continue to devour their host galaxy until nothing is left. Blackholes grow according to a inverse logarithmic scale. The larger they are, the more mass it takes to grow. Eventually(rather quickly actually), blackholes run out of nearby stars they can influence with their gravitational pull. That is btw, why quasars can only be observed in very far away galaxies: because we're actually observing galaxies when they were very very young and quasars were still possible. $\endgroup$ – AngelPray Jan 7 '17 at 6:38
1
$\begingroup$

Two huge problems:

1) Quasars existed in early universe, when there was very little elements outside hydrogen and helium - thus forming a rocky planet was rather hard (@JDługosz already pointed that out)

2) Quasars are terribly variable "stars" - a black hole devouring nearby gas in wholesale quantities. So do not expect any orbit to be in Goldilocks zone for long.

$\endgroup$
  • $\begingroup$ This should be a comment, not posted as an answer. Like the one I already left covering your first point. $\endgroup$ – JDługosz Jan 9 '17 at 8:17
  • $\begingroup$ Quasar activity peaked at redshifts $z\sim2\text{-}3$, i.e. when the universe was a few billion years old. Many generations of stars had already lived and died, meaning that there should indeed be plenty of heavy elements. Some of those heavy elements - double-ionized oxygen, for instance - are abundant enough to be used to probe quasars and other active galactic nuclei. $\endgroup$ – HDE 226868 Apr 2 '19 at 13:39
1
$\begingroup$

AGN structure and emission

Let's talk about the structure of an active galactic nucleus like a quasar, and the types of emission we see from it. The classic unified model of an AGN consists of a supermassive black hole (of perhaps $\sim10^8\text{-}10^9M_{\odot}$) surrounded by an accretion disk about $\sim10^{13}\text{-}10^{14}$ meters in radius. The disk has a radial temperature distribution of $$T(r)\approx3\times10^5\dot{m}\left(\frac{M}{10^8M_{\odot}}\right)^{1/4}\left(\frac{r}{r_{\text{Sch}}}\right)^{-3/4}\text{ K}$$ where $r_{\text{Sch}}$ is the Schwarzschild radius and $\dot{m}$ is the ratio of the accretion rate to the maximum accretion rate specified by the Eddington limit. If we assume that $\dot{m}\approx1$ and $M\sim10^8M_{\odot}$, the luminosity should be about $L\sim10^{39}\text{ W}$. The disk itself should emit most strongly in the x-ray and ultraviolet portion of the spectrum, with UV emission beginning at $\sim10^{13}\text{ m}$.

Beyond the disk lies the broad-line region, where high-velocity gas clouds produce secondary emission. This area should have an outer radius of maybe $\sim10^{14}\text{-}10^{15}$ meters. Beyond the broad-line region lies the narrow-line region, (radius $\sim1000$ light-years) which includes the obscuring torus (radius $\sim100$ light-years), the latter being a structure of gas and dust that may be fed by a wind from the accretion disk. The narrow-line region contains slower-moving gas clouds; the low speeds produces less Doppler broadening - hence the name.

I think that what you've been considering is only the jets that arise from the accretion disk. Matter from the disk travels along magnetic field lines; electrons are accelerated, producing the synchotron emission observed from many AGN. This is indeed strong, but keep in mind that the jets are narrow and usually perpendicular to the plane of the galaxy, meaning that most objects in the galaxy are far away from the jet. If your planet exists in the equatorial plane of the quasar, it won't be hit by the jets, although it could experience radiation from the accretion disk.

Different radii

We could attempt to compute the flux the planet would receive from the disk via the inverse square law if the disk was a point source and emitting isotropically. This is decidedly not the case. If we want to look at the best-case scenario, where the black hole is accreting below the Eddington limit, we could attempt to model the disk as a thin disk and use the Shakura-Sunyaev model, where the flux is given by $$F(r)=\frac{3GM\dot{M}}{8\pi r^3}\left(1-\sqrt{\frac{r_0}{r}}\right)$$ where $r_0$ is the inner radius of the disk. If we assume the accretion disk somehow extends to the black hole's surface, we get that at a distance of $1.17\times10^{16}$ meters (1.24 light-years), the flux from the disk is about 10% of that from the Sun.

There's another radius we may want to consider, which is $r_{\text{blr}}$, the outer radius of the broad-line region. It's calculated by $$r_{\text{blr}}\approx0.26\times10^{15}\left(\frac{L}{10^{37}\text{ W}}\right)^{1/2}\text{ m}\approx0.27\text{ light-years}$$ This radius is quite similar to the dust sublimation radius, inside which dust will be vaporized. This occurs when $T\approx1500\text{ K}$. I then calculate that for our supermassive black hole, $r_{\text{sub}}\approx0.037$ light-years. The order-of-magnitude difference is because of my assumption that the black hole is accreting as fast as it can - a sort of worst-case scenario for our quasar. Finally, consider that the largest accretion disks may be around $0.01$ light-years in radius.

Here's a summary of the various length scales:

  • $2\text{ AU}$: the Schwarzschild radius of the black hole
  • $0.001\text{ ly}$ ($64\text{ AU}$): the outermost point of UV emission
  • $0.01\text{ ly}$: the outer radius of the accretion disk
  • $0.037\text{ ly}$: the dust sublimation radius
  • $0.27\text{ ly}$: the outer radius of the broad-line region
  • $1.24\text{ ly}$: the distance at which the jet's flux becomes 10% of that from the Sun
  • $\sim100\text{ ly}$: the outer edge of the obscuring torus
  • $\sim1000\text{ ly}$: the outer edge of the narrow-line region

The answer to your question, then, depends on the innermost zone you're comfortable with having your planets in, assuming you're good with the systems being in the plane. If not, you can be around 80 light-years away, for a direct hit from the jet.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.